Roboguru

Pada persegi panjang ,  dan . Jika  dan , maka

Pertanyaan

Pada persegi panjang text ABCD end texttext AB=12 end text dan text BC=5 end text. Jika text AB= end text u with rightwards arrow on top dan text AC= end text v with rightwards arrow on top, maka u with rightwards arrow on top times v with rightwards arrow on top equals space horizontal ellipsis 

  1. 13 

  2. 60 

  3. 65 

  4. 120 

  5. 144 

Pembahasan Soal:

Panjang AC ditentukan dengan rumus pythagoras berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell text AC end text end cell equals cell square root of text AB end text squared plus text BC end text squared end root end cell row blank equals cell square root of 12 squared plus 5 squared end root end cell row blank equals cell square root of 144 plus 25 end root end cell row blank equals cell square root of 169 end cell row blank equals 13 end table

Pada segitiga siku-siku ABC, text cos A end text dapat ditentukan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell text cos A end text end cell equals cell fraction numerator text AB end text over denominator text AC end text end fraction end cell row blank equals cell 12 over 13 end cell end table

Perkalian vektor (dot product) dapat ditentukan sebagai berikut.

u with rightwards arrow on top times v with rightwards arrow on top equals open vertical bar u with rightwards arrow on top close vertical bar times open vertical bar v with rightwards arrow on top close vertical bar times cos space theta

Penyelesaian perkalian vektor tersebut adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell u with rightwards arrow on top times v with rightwards arrow on top end cell equals cell open vertical bar u with rightwards arrow on top close vertical bar times open vertical bar v with rightwards arrow on top close vertical bar times cos space theta end cell row blank equals cell 12 times 13 times 12 over 13 end cell row blank equals 144 end table

Oleh karena itu, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Eka

Mahasiswa/Alumni Universitas Pendidikan Indonesia

Terakhir diupdate 06 Juni 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Diketahui , dan . Besar sudut antara vektor  dan  adalah ....

Pembahasan Soal:

Diketahui open vertical bar straight a with rightwards arrow on top close vertical bar equals square root of 6 comma space open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 0, dan straight a with rightwards arrow on top times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals 3. Maka:

- Menentukan panjang straight b with rightwards arrow on top 

space space space space space space space space space space space space space space open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses times open parentheses straight a with rightwards arrow on top plus straight b with rightwards arrow on top close parentheses equals 0 straight a with rightwards arrow on top times straight a with rightwards arrow on top plus straight a with rightwards arrow on top times straight b with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top minus straight b with rightwards arrow on top times straight b with rightwards arrow on top equals 0 space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space open parentheses square root of 6 close parentheses squared minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space 6 minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space minus open vertical bar straight b with rightwards arrow on top close vertical bar squared equals negative 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar squared equals 6 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space open vertical bar straight b with rightwards arrow on top close vertical bar equals plus-or-minus square root of 6 ,

- Menentukan straight a with rightwards arrow on top times straight b with rightwards arrow on top 

space space space space space space straight a with rightwards arrow on top times open parentheses straight a with rightwards arrow on top minus straight b with rightwards arrow on top close parentheses equals 3 space space space space straight a with rightwards arrow on top times straight a with rightwards arrow on top minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 space space space space open vertical bar straight a with rightwards arrow on top close vertical bar squared minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 open parentheses square root of 6 close parentheses squared minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 space space space space space space space space space space minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 minus 6 space space space space space space space space space space minus straight a with rightwards arrow on top times straight b with rightwards arrow on top equals negative 3 space space space space space space space space space space space space space straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

Sehingga besar sudut antara vektor straight a with rightwards arrow on top dan straight b with rightwards arrow on top adalah:

- Untuk open vertical bar straight b with rightwards arrow on top close vertical bar equals square root of 6 dan straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar times open vertical bar straight b with rightwards arrow on top close vertical bar end fraction cos space straight theta equals fraction numerator 3 over denominator square root of 6 times square root of 6 end fraction cos space straight theta equals 3 over 6 cos space straight theta equals 1 half cos space straight theta equals cos space 30 degree space space space space space space space straight theta equals 30 degree space space space space space space space straight theta equals 30 degree cross times fraction numerator straight pi over denominator 180 degree end fraction space space space space space space space space straight theta equals straight pi over 6 

- Untuk open vertical bar straight b with rightwards arrow on top close vertical bar equals negative square root of 6 dan straight a with rightwards arrow on top times straight b with rightwards arrow on top equals 3 

cos space straight theta equals fraction numerator straight a with rightwards arrow on top times straight b with rightwards arrow on top over denominator open vertical bar straight a with rightwards arrow on top close vertical bar times open vertical bar straight b with rightwards arrow on top close vertical bar end fraction cos space straight theta equals fraction numerator 3 over denominator square root of 6 times negative square root of 6 end fraction cos space straight theta equals fraction numerator 3 over denominator negative 6 end fraction cos space straight theta equals negative 1 half cos space straight theta equals cos space 120 degree space space space space space space space straight theta equals 120 degree space space space space space space space straight theta equals 120 degree cross times fraction numerator straight pi over denominator 180 degree end fraction space space space space space space space space straight theta equals 2 over 3 straight pi 

Jadi, jawaban yang benar adalah A dan E.

Roboguru

Jika vektor  dan vektor  membentuk sudut , panjang vektor  dan panjang vektor , maka  sama dengan ...

Pembahasan Soal:

Perhatikan gambar berikut.



 

Perkalian titik (dot product) antara a with rightwards arrow on top dan b with rightwards arrow on top dapat dirumuskan sebagai berikut:

a with rightwards arrow on top times b with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space theta 

dimana

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals cell panjang space vektor space a with rightwards arrow on top end cell row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals cell panjang space vektor space b with rightwards arrow on top end cell end table  

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell open vertical bar a with rightwards arrow on top close vertical bar end cell equals 4 row cell open vertical bar b with rightwards arrow on top close vertical bar end cell equals 6 row theta equals cell 120 degree end cell end table 

Karena a with rightwards arrow on top times a with rightwards arrow on top equals open vertical bar a with rightwards arrow on top close vertical bar squared, maka nilai a with rightwards arrow on top times open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses adalah:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with rightwards arrow on top times open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses end cell equals cell a with rightwards arrow on top times b with rightwards arrow on top plus a with rightwards arrow on top times a with rightwards arrow on top end cell row blank equals cell open vertical bar a with rightwards arrow on top close vertical bar open vertical bar b with rightwards arrow on top close vertical bar space cos space theta plus open vertical bar a with rightwards arrow on top close vertical bar squared end cell row blank equals cell 4 times 6 times cos space 120 degree plus 4 squared end cell row blank equals cell 24 times cos space left parenthesis 180 degree minus 120 degree right parenthesis plus 16 end cell row blank equals cell 24 times open parentheses negative cos space 60 degree close parentheses plus 16 end cell row blank equals cell 24 times open parentheses negative 1 half close parentheses plus 16 end cell row blank equals cell negative 12 plus 16 end cell row blank equals 4 end table 

Sehingga, table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell a with rightwards arrow on top end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank times end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell open parentheses b with rightwards arrow on top plus a with rightwards arrow on top close parentheses end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank equals blank end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank 4 end table.

Jadi, jawaban yang tepat adalah A.

Roboguru

Diketahui  dan . Hasil dari  adalah ....

Pembahasan Soal:

Diketahui open vertical bar a with bar on top close vertical bar equals 4 comma space open vertical bar b with bar on top close vertical bar equals 3 dan angle open parentheses a with bar on top comma space b with bar on top close parentheses equals 45 degree, maka didapatkan:

table attributes columnalign right center left columnspacing 0px end attributes row cell a with bar on top times b with bar on top end cell equals cell open vertical bar a with bar on top close vertical bar times open vertical bar b with bar on top close vertical bar times cos space 45 degree end cell row blank equals cell 4 times 3 times 1 half square root of 2 end cell row blank equals cell 6 square root of 2 end cell end table 

Dengan demikian, hasil dari a with bar on top times b with bar on top adalah 6 square root of 2.

Jadi, jawaban yang benar adalah C.

Roboguru

Besar sudut antara  adalah ....

Pembahasan Soal:

Diketahui a with rightwards arrow on top equals open parentheses table row 3 row 2 row 4 end table close parentheses space dan space b with rightwards arrow on top equals open parentheses table row 2 row 3 row cell negative 3 end cell end table close parentheses.

Perhatikan:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator a with rightwards arrow on top times b with rightwards arrow on top over denominator open vertical bar a with rightwards arrow on top close vertical bar times open vertical bar b with rightwards arrow on top close vertical bar end fraction end cell end table 

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space alpha end cell equals cell fraction numerator 3 cross times 2 plus 2 cross times 3 plus 4 cross times left parenthesis negative 3 right parenthesis over denominator square root of 3 squared plus 2 squared plus 4 squared end root cross times square root of 2 squared plus 3 squared plus left parenthesis negative 3 right parenthesis squared end root end fraction end cell row blank equals cell fraction numerator 6 plus 6 minus 12 over denominator square root of 29 cross times square root of 22 end fraction end cell row cell cos space alpha end cell equals cell fraction numerator 0 over denominator square root of 29 cross times square root of 22 end fraction end cell row cell cos space alpha end cell equals 0 row alpha equals cell 90 degree space atau space alpha equals 270 degree end cell end table

Jadi, besar sudut antara a with rightwards arrow on top space dan space b with rightwards arrow on top adalah 90 degree space atau space 270 degree.

Roboguru

Jika  dan  maka besar sudut yang dibentuk oleh vektor  dan  sama dengan...

Pembahasan Soal:

Ingat kembali  penjumlahan vektor, panjang vektor dan besar sudut pada vektor berikut.

  • Jika top enclose a equals a subscript 1 i plus a subscript 2 j plus a subscript 3 k dan top enclose b equals b subscript 1 i plus b subscript 2 j plus b subscript 3 k maka

top enclose a plus top enclose b equals left parenthesis a subscript 1 plus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 plus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 plus b subscript 3 right parenthesis k

top enclose a minus top enclose b equals left parenthesis a subscript 1 minus b subscript 1 right parenthesis i plus left parenthesis a subscript 2 minus b subscript 2 right parenthesis j plus left parenthesis a subscript 3 minus b subscript 3 right parenthesis k 

  • Jika theta adalah sudut antara vektor top enclose a space dan space top enclose b, maka cos space theta equals fraction numerator a bullet b over denominator open vertical bar top enclose a close vertical bar times open vertical bar top enclose b close vertical bar end fraction.

 

  • open vertical bar top enclose a close vertical bar equals square root of a subscript 1 squared plus a subscript 2 squared plus a subscript 3 squared end root 

 

  • a bullet b equals a subscript 1 b subscript 1 plus a subscript 2 b subscript 2 plus a subscript 3 b subscript 3

Dari rumus di atas, diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis top enclose a plus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 plus 1 right parenthesis i plus left parenthesis 1 plus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 plus 2 right parenthesis k end cell row blank equals cell 0 i plus 0 j plus 4 k end cell row cell left parenthesis top enclose a minus top enclose b right parenthesis end cell equals cell left parenthesis negative 1 minus 1 right parenthesis i plus left parenthesis 1 minus left parenthesis negative 1 right parenthesis right parenthesis j plus left parenthesis 2 minus 2 right parenthesis k end cell row blank equals cell negative 2 i plus 2 j plus 0 k end cell row blank blank blank end table 

Sehingga besar sudut  yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis adalah

table attributes columnalign right center left columnspacing 0px end attributes row cell cos space theta end cell equals cell fraction numerator left parenthesis top enclose a plus top enclose b right parenthesis bullet left parenthesis top enclose a minus top enclose b right parenthesis over denominator open vertical bar top enclose a plus top enclose b close vertical bar times open vertical bar top enclose a minus top enclose b close vertical bar end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 i plus 0 j plus 4 k right parenthesis bullet left parenthesis negative 2 i plus 2 j plus 0 k right parenthesis over denominator square root of 0 squared plus 0 squared plus 4 squared end root cross times square root of left parenthesis negative 2 right parenthesis squared plus 2 squared plus 0 squared end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator left parenthesis 0 cross times 0 right parenthesis plus left parenthesis 0 cross times 2 right parenthesis plus left parenthesis 4 cross times 0 right parenthesis over denominator square root of 0 plus 0 plus 16 end root cross times square root of 4 plus 4 plus end root end fraction end cell row cell cos space theta end cell equals cell fraction numerator 0 over denominator square root of 16 cross times square root of 8 end fraction end cell row cell cos space theta end cell equals 0 row cell cos space theta end cell equals cell cos space 90 degree end cell row theta equals cell 90 degree end cell end table 

Jadi, besar sudut yang dibentuk oleh vektor left parenthesis top enclose a plus top enclose b right parenthesis dan left parenthesis top enclose a minus top enclose b right parenthesis  adalah 90 degree.

Oleh karena itu, jawaban yang benar adalah D.

 

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

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